Hybridization states of Ni atom in [NiCl2(PPh3)2], [Ni(HDMG)2], and [Ni(CO)4] complexes are as follows:

1. [NiCl2(PPh3)2]:
The nickel atom in this complex is surrounded by two chloride ions and two triphenylphosphine ligands. The electronic configuration of nickel is [Ar] 3d8 4s2. To determine the hybridization state, we need to consider the number of sigma bonds and lone pairs around the central atom. In this case, nickel forms two sigma bonds with the chloride ions and two sigma bonds with the phosphorus atoms. There are no lone pairs on the nickel atom. Therefore, the hybridization state of the nickel atom is sp3.

2. [Ni(HDMG)2]:
In this complex, the nickel atom is coordinated to two HDMG (hexadentate dimethylglyoximate) ligands. The electronic configuration of nickel is [Ar] 3d8 4s2. The HDMG ligands coordinate to the nickel atom through the oxygen atoms of the carbonyl groups and the nitrogen atoms of the oxime groups. The nickel atom forms four sigma bonds with the HDMG ligands and has no lone pairs. Therefore, the hybridization state of the nickel atom is dsp2.

3. [Ni(CO)4]:
In this complex, the nickel atom is coordinated to four carbon monoxide ligands. The electronic configuration of nickel is [Ar] 3d8 4s2. The carbon monoxide ligands coordinate to the nickel atom through the carbon atom. Each carbon monoxide ligand forms a sigma bond with the nickel atom using its lone pair of electrons. There are no lone pairs on the nickel atom. Therefore, the hybridization state of the nickel atom is sp3.

In summary, the hybridization states of the nickel atom in the given complexes are:
a) [NiCl2(PPh3)2]: sp3
b) [Ni(HDMG)2]: dsp2
c) [Ni(CO)4]: sp3

Note: The given answer is option 'B' which corresponds to the correct hybridization states of the nickel atom in the three complexes.